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Stolb23 [73]
3 years ago
14

A leopard with a mass of 55.00 kg climbs 12.0 m up a tree. What is it’s gain in GPE

Physics
1 answer:
PSYCHO15rus [73]3 years ago
4 0
The answer is: 6,468 j
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If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit
marissa [1.9K]

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})

where

E_0 = 13.6 eV is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV

Converting into Joules,

E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J

And now we can find the wavelength of the emitted photon by using the equation

E=\frac{hc}{\lambda}

where h is the Planck constant and c is the speed of light. Solving for \lambda,

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm

And this wavelength corresponds to red light.

5 0
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